Áp dụng BĐT Bunhiacopxki , ta có: 

Với a,b,c,d >0

\(\left(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\right)\left[a\left(b+c\right)+b\left(c+d\right)+c\left(d+a\right)+d\left(a+b\right)\right]\ge\left(a+b+c+d\right)^2\)

\(\Rightarrow\left(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\right)\ge\frac{\left(a+b+c+d\right)^2}{ab+bc+cd+da+2ca+2bd}\)

Ta cần chứng minh : 

\(\left(a+b+c+d\right)^2\ge2\left(ab+bc+cd+da+2ac+2bd\right)\)

\(\Leftrightarrow a^2+b^2+c^2+d^2\ge2ca+2bd\)

\(\Leftrightarrow\left(a-c\right)^2+\left(b-d\right)^2\ge0\)(đúng) 

\(\Leftrightarrow dpcm\)

Áp dụng bđt Cô-si: \(a^2+b^2+c^2+d^2\)\(\ge4\sqrt[4]{a^2.b^2.c^2.d^2}\)\(=4\sqrt[4]{\left(abcd\right)^2}=4\sqrt[4]{1^2}=4;\)

\(a\left(b+c\right)+b\left(c+d\right)+d\left(c+a\right)=ab+ac+bc+bd+dc+da\)

\(\ge6\sqrt[6]{ab.ac.bc.bd.dc.da}=6\sqrt[6]{\left(abcd\right)^3}=6\sqrt[6]{1^3}=6\)

=>\(a^2+b^2+c^2+d^2\)\(a\left(b+c\right)+b\left(c+d\right)+d\left(c+a\right)\ge4+6=10\)

Dấu "=" xảy ra khi a=b=c=d=1

\(VT=\frac{a+b-\left(b+d\right)}{d+b}+\frac{\left(d+c\right)-\left(b+c\right)}{b+c}+\frac{\left(b+a\right)-\left(a+c\right)}{c+a}+\frac{\left(c+d\right)-\left(a+d\right)}{a+d}\)

\(VT=\frac{a+b}{d+b}-1+\frac{\left(d+c\right)}{b+c}-1+\frac{\left(b+a\right)}{c+a}-1+\frac{\left(c+d\right)}{a+d}-1\)

\(VT=\left(a+b\right).\left(\frac{1}{d+b}+\frac{1}{a+c}\right)+\left(d+c\right).\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\)

Chứng minh đc bđt sau: Với x; y > 0 ta có  \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Áp dụng ta có: \(VT\ge\left(a+b\right).\frac{4}{d+b+a+c}+\left(d+c\right).\frac{4}{b+c+a+d}-4\ge\frac{4.\left(a+b+c+d\right)}{a+b+c+d}-4=0\)

=> ĐPCM

Cộng 4 vào vế trái nhá

\(VT+4=\left(\dfrac{a-d}{d+b}+1\right)+\left(\dfrac{d-b}{b+c}+1\right)+\left(\dfrac{b-c}{c+a}+1\right)+\left(\dfrac{c-a}{a+d}+1\right)\)

\(=\dfrac{a+b}{d+b}+\dfrac{d+c}{b+c}+\dfrac{a+b}{c+a}+\dfrac{c+d}{a+d}\)

\(=\left(a+b\right)\left(\dfrac{1}{d+b}+\dfrac{1}{c+a}\right)+\left(c+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)\)

\(\ge\left(a+b\right).\dfrac{4}{a+b+c+d}+\left(c+d\right).\dfrac{4}{a+b+c+d}\)

\(=\left(a+b+c+d\right).\dfrac{4}{a+b+c+d}\)\(=4\)

\(\Rightarrow VT\ge0=VP\)(Đpcm)

Cố gắng giúp mik nhé.  Mik đang ôn thi

bạn ơi bạn làm dc chưa

abcd = 1 \(\Rightarrow\hept{\begin{cases}ab=\frac{1}{cd}\\ac=\frac{1}{bd}\\bc=\frac{1}{ad}\end{cases}}\)

Áp dụng bđt AM-GM ta có:

A = \(a^2+b^2+c^2+d^2+a\left(b+c\right)+b\left(c+d\right)+d\left(c+a\right)\)\(=\left(a^2+b^2+ab\right)+\left(c^2+d^2+cd\right)+ac+bc+bd+ad\)

\(=\left(a^2+b^2+ab\right)+\left(c^2+d^2+cd\right)+\left(\frac{1}{bd}+bd\right)+\left(\frac{1}{ad}+ad\right)\)

\(\ge3\sqrt{a^2.b^2.ab}+3\sqrt{c^2.d^2.cd}+2\sqrt{\frac{1}{bd}.bd}+2\sqrt{\frac{1}{ad}.ad}\)

\(\Leftrightarrow A\ge3ab+3cd+2+2\)\(=\frac{3}{cd}+3cd+4\ge2\sqrt{\frac{3}{cd}.3cd}+4=6+4=10\)

Dấu "=" xảy ra khi a = b = c = d = 1

cố gắng giúp mình nha

\(\frac{a}{b}<\frac{c}{d}\) \(\Rightarrow\) ad < bc.

\(\Rightarrow\) ad + ab < bc + ab

       a.(d + b) < b(a+c) 

\(\Rightarrow\) \(\frac{a}{b}<\frac{a+c}{b+d}\)  (1)

ad< bc         \(\Rightarrow\) ad + cd < bc + cd

                        d.(a+c) < c(b+d)

                    \(\Rightarrow\frac{a+c}{b+d}<\frac{c}{d}\)  (2)

Từ (1) và (2) suy ra: \(\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}\)